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A body of mass ' m ' dropped from a height ' h ' reaches the ground with a speed of 0.8 \sqrt{g h}. The value of workdone by the air-friction is :
 
Option: 1 -0.68 \mathrm{mgh}
Option: 2 mgh
Option: 3 0.64 Mgh
Option: 4 1.64 mgh

Answers (1)

best_answer


v= 0\cdot 8\sqrt{gh}
In absence of air friction, the speed would have been v_{0}= \sqrt{2gh}
KE_{lost}= W_{friction}
i.e, KE lost by body is work done by friction
\left ( \frac{1}{2}mv_{0}^{2}-\frac{1}{2}mv^{2} \right )= \left | W_{friction} \right |

\left ( \frac{1}{2}mv_{0}^{2}-\frac{1}{2}mv^{2} \right )= \left | W_{friction} \right |
\frac{1}{2}m\left ( 2gh-0\cdot 64gh \right )= \left | W_{friction} \right |
-0\cdot 68\: mgh= W_{friction}

Negative sign implies here work done by frictional force is opposite to displacement. The correct option is (1)

Posted by

vishal kumar

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